squaring complex numbers with radicals\nscore: 0/1 penalty: 1 off\nquestion\nexpress (\\frac{\\sqrt{3}}{2}-\\…

Explanation:

Step1: Apply the formula $(a - b)^2=a^{2}-2ab + b^{2}$

Let $a=\frac{\sqrt{3}}{2}$ and $b = \frac{3}{4}i$. Then $(\frac{\sqrt{3}}{2}-\frac{3}{4}i)^{2}=(\frac{\sqrt{3}}{2})^{2}-2\times\frac{\sqrt{3}}{2}\times\frac{3}{4}i+(\frac{3}{4}i)^{2}$.

Step2: Calculate each term

$(\frac{\sqrt{3}}{2})^{2}=\frac{3}{4}$, $2\times\frac{\sqrt{3}}{2}\times\frac{3}{4}i=\frac{3\sqrt{3}}{4}i$, and $(\frac{3}{4}i)^{2}=\frac{9}{16}i^{2}$. Since $i^{2}=-1$, then $(\frac{3}{4}i)^{2}=-\frac{9}{16}$.

Step3: Combine the terms

$\frac{3}{4}-\frac{3\sqrt{3}}{4}i-\frac{9}{16}=\frac{12 - 9}{16}-\frac{3\sqrt{3}}{4}i=\frac{3}{16}-\frac{3\sqrt{3}}{4}i$.

Answer:

$\frac{3}{16}-\frac{3\sqrt{3}}{4}i$